If the volume of a parallelopiped,

We know that the volume of parallelopiped

$=\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$

$\left|\begin{array}{ccc}1 & 1 & n \\ 2 & 4 & -n \\ 1 & n & 3\end{array}\right|=158$

$\Rightarrow\left(12+n^{2}\right)-1(6+n)+n(2 n-4)=158$

$\Rightarrow 3 n^{2}-5 n-152=0$

$\Rightarrow 3 n^{2}-24 n+19 n-152=0$

$\Rightarrow 3 n(n-8)+19(n-8)=0$

$\Rightarrow n=8$ or $n=\frac{-19}{3}$

$\therefore \bar{a}=\hat{i}+\hat{j}+8 \hat{k}, \bar{b}=2 \hat{i}+4 \hat{j}-8 \hat{k}$ and

$\bar{c}=\hat{i}+8 \hat{j}+3 \hat{k}$

$\bar{a} \cdot \bar{c}=1+8+24=33$

$\bar{b} \cdot \bar{c}=2+32-24=10$